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If you have an elastic collision between objects 1 and 2 and where 'kinetic energy is conserved' ... will both the objects always 'join' and have the same common velocity? Linear momentum will be conserved, so if the objects 'join' $$\vec{p}_1+\vec{p}_2=(m_1+m_2)\vec{V}\ $$ where $\vec{V}$ is the final common velocity. The initial kinetic energy is ...


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The force for a realistic collision will not be constant over the course of the collision. There is no one way to figure out what the force on each body will be over time as the collision progresses. The one thing we can say for sure is that $$\Delta \vec{p} = \int_{t_0}^{t_1}\vec{F}(t)dt$$ where $\Delta \vec{p}$ is the change in momentum of one of the ...


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As Jack M has answered, if an object of mass $m$ is having an acceleration of $a$ this means that the net force $F$ acting on the object is $ma$. In this question the basketball of mass $m$ is having a downward acceleration of $g$ due to earth's gravity. The force acting on the basketball is $mg$. We have assumed here that everything is happening on the ...


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You're not stupid. You've shown you have intellectual curiosity about extremely complicated topics you've only just been introduced to. "1.If force equals mass times acceleration, wouldn't a basketball dropped from the top of the Eiffel tower exert the same force on the ground as a basketball dropped a foot off the ground? They both have the same mass, and ...


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Force does equal mass times acceleration. However $9.8~\text{m/s^2}$ is the acceleration of the ball imposed by gravity. The acceleration that the ball experiences upon impact with the ground is instead proportional to its current velocity. Upon impact, $a=-\frac{v}{t}$, where v is current velocity and t is the time impact lasts. If the ball were ...


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I believe you are confusing force with energy -- kinetic energy, to be more precise. Remember $$E {\scriptsize _K} = \textstyle \frac{1}{2} m v^2$$ The ball dropped from the higher location falls a greater distance with nothing opposing the force of gravity. Therefore, the ball accelerates to a higher velocity than the ball dropped from 1 ft. Using ...


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I'll set up the problem in the following way. Let's assume the line of centers is along the x-axis. Initial speed of $m_1$ is $u_1$ at an angle $\theta$ to the x-axis. Final speed of $m_{1}$ is $v_1$ at an angle $\alpha$ to the x-axis. The final speed of $m_2$ is $v_2$ and is along the line of centers (the x-axis), such that $\alpha$ is the angle between the ...


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I would reason along these lines. Put the system in the centre of mass. Then, at collision, consider the plane through the centre of mass and perpendicular to the line through the centres of the balls. Since this plane is fixed in the frame of the centre of mass, this behaves like a wall the balls smash against, and therefore the velocity vectors after ...


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For this post I'm using the subscripts $_0$, $_1$, $_p$, and $_n$ to denote pre-impact, post-impact, normal to the impact surface, and tangential to the impact surface as per the OP's diagrams. Set Up If the ball is spherical and of uniform density, $I=\frac25m\,r^2$ The force acting on the ball over time can be integrated into an impulse. The impulse ...


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What will be the difference between the force acting on the top and that on bottom of the box? Is it equal to the weight of the ball? So will the difference in the force acting on the top and that on the bottom of the box be amplified The rebound of the ball can never be amplified. Even in an idealized example in which the collision is perfectly ...


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Simple answer: you have to use the information you are given. I am not quite sure what equation of motion you intend to use - you have very little information about the interaction at the floor, only that the object apparently lost some energy (which is why it ends up reaching a lower height after the bounce). The simplest way to do this is with ...


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You can use either method, conservation of energy or kinematics, and both will give the same answer: CONSERVATION OF ENERGY Let's arbitrarily define downwards displacements, velocities, accelerations and forces as negative. Falling motion: $$mgh_1 = \frac{1}{2}mv_1^2$$ $$\therefore v_1 = -\sqrt{2gh_1}$$ Rising motion: $$\frac{1}{2}mv_2^2 = mgh_2$$ ...


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If the point of the air track is to eliminate friction from consideration, let's make the situation even simpler: Imagine a completely empty universe. Total blackness, no objects, no light, nothing. Now add two clay lumps, and an invisible observer (you). One clay lump appears (to you) to be motionless. Another clay lump appears (to you) to be moving ...


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This is a pretty long answer - skip to the bottom if you just want to see the solution. It would appear that you are assuming the collision is sufficiently inelastic that the ball does not bounce off the surface, correct? Seems a bit unusual for a ball to be inelastic enough to stick to a surface, but stiff enough to roll smoothly afterwards, but we'll go ...


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They must necessarily deform the same amount, otherwise you would violate Galilean invariance -- the idea that physics works no matter what reference frame you are in. Suppose the stationary object $A$ deformed more. While you sit still in $A$'s reference frame, watching it deform more, have your friend run alongside the moving object $B$. In your friend's ...


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The reference frame doesn't matter. You cannot define a reference frame the way you did. Their deformation should be the same. The reason for this is that you can always transform to a different reference frame, i.e the centre of mass frame and physics should be the same as in any other inertial frame. Having established this, in the centre of mass frame ...



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